Matrices And Determinants – JEE Main MathematicsPractice Questions & PYQs

The determinant of a matrix scaled by a constant factor $k$ follows the property $|kA| = k^n|A|$, where $n$ represents the order of the square matrix. Since $A$ is a $3 \times 3$ matrix, the determinant of $2A$ is $2^3|A|$, which simplifies to $8|A|$. Substituting this into the given equation $|2A|^3 = 2^{21}$ leads to $(8|A|)^3 = 2^{21}$, or $(2^3|A|)^3 = 2^{21}$, which simplifies further to $2^9|A|^3 = 2^{21}$. Dividing both sides by $2^9$ reveals that $|A|^3 = 2^{12}$, meaning $|A| = 2^4 = 16$. By calculating the determinant of the matrix $A$ along the first row, we obtain $|A| = 1(\alpha^2 - \beta^2) = \alpha^2 - \beta^2$. Therefore, we must satisfy the integer equation $\alpha^2 - \beta^2 = 16$, which can be factored into $(\alpha - \beta)(\alpha + \beta) = 16$. Considering pairs of factors that yield $16$ while ensuring $\alpha$ and $\beta$ remain integers, setting $\alpha - \beta = 2$ and $\alpha + \beta = 8$ produces the system $2\alpha = 10$, resulting in $\alpha = 5$.

Expanding the determinant along the third row allows for a straightforward evaluation, resulting in the expression $(2\alpha+3)\left(\frac{7\alpha}{6}\right) - (3\alpha+1)\left(-\frac{7}{6}\right) = 0$. Dividing the entire equation by the common factor $\frac{7}{6}$ leads to the simplified quadratic equation $2\alpha^{2+}6\alpha+1=0$. Using the quadratic formula, the values are $\alpha = \frac{-3 \pm \sqrt{7}}{2}$. Since $\sqrt{7}$ is approximately $2.65$, both resultant values of $\alpha$ are approximately $-0.175$ and $-2.825$, which fall within the interval $(-3, 0)$.

The behavior of a system of linear equations is primarily dictated by its coefficient matrix determinant, which for this system evaluates to $\Delta = (2\lambda - 1)^2$. As long as $\lambda \neq \frac{1}{2}$, the determinant remains non-zero, which guarantees that the system possesses a unique, consistent solution regardless of the value of $\mu$.

When $\lambda = \frac{1}{2}$, the determinant becomes zero, and the system either lacks a solution or yields infinitely many solutions depending on the value of $\mu$. Evaluating the conditions for consistency reveals that infinite solutions occur specifically when $\mu = 1$ or $\mu = 15$, meaning the system is only inconsistent when $\lambda = \frac{1}{2}$ and $\mu$ is neither of these values. Consequently, the assertion that the system is inconsistent whenever $\lambda = \frac{1}{2}$ and $\mu \neq 1$ is inaccurate, as it incorrectly characterizes the condition where $\mu = 15$ as inconsistent, when the system actually admits infinite solutions in that case.

Understanding the power structure of the adjugate matrix, where

, allows us to compute $B$ by summing these matrices from $k=0$ to $10$. Summing the elements across all eleven matrices, the diagonal entries each become $11$, while the top-right entry becomes $-2 \times \frac{10 \times 11}{2}$, which is $-110$. With the bottom-left entry remaining $0$, the matrix $B$ equals

, and summing all its components yields $-88$.

The Cayley-Hamilton theorem, which states that every square matrix satisfies its own characteristic equation, provides the most efficient path to the answer. By simplifying the given matrix expressions involving $A$, $B$, and $C$, we find that the matrix $C^2$ is equivalent to

. Since $C^2 = B$, the characteristic equation for $C^2$ is determined by calculating $\det(B - \lambda I) = 0$.

Expanding the determinant

results in $(1-\lambda)(5-\lambda) - 3 = 0$, which simplifies to $\lambda^2 - 6\lambda + 5 - 3 = 0$, or $\lambda^2 - 6\lambda + 2 = 0$. Applying the Cayley-Hamilton theorem by substituting the matrix $C^2$ for $\lambda$, we obtain the matrix equation $C^4 - 6C^2 + 2I = O$. Comparing this result with the given expression $C^4 + \alpha C^2 + \beta I = O$, we identify the values $\alpha = -6$ and $\beta = 2$. Calculating the final expression $2\beta - \alpha$ yields $2(2) - (-6)$, which equals $10$.

A system of linear equations possesses infinitely many solutions when the planes defined by these equations are linearly dependent, meaning they intersect along a common line. To identify the parameters that satisfy this condition, we can express the first equation as a linear combination of the other two, represented by constants $k_1$ and $k_2$ such that $2x + 3y - z - 5 = k_1(x + \alpha y + 3z + 4) + k_2(3x - y + \beta z - 7)$. Equating the corresponding coefficients of $x, y, z,$ and the constant terms on both sides yields the system $k_1 + 3k_2 = 2$, $k_1\alpha - k_2 = 3$, $3k_1 + \beta k_2 = -1$, and $4k_1 - 7k_2 = -5$. Solving the two equations involving only $k_1$ and $k_2$ gives $k_2 = 13/19$ and $k_1 = -1/19$. Substituting these specific values into the equations for the variable coefficients allows us to determine $\alpha = -70$ and $\beta = -16/13$. Computing the final product $13\alpha\beta$ involves calculating $13(-70)(-16/13)$, which simplifies to 1120.

The determinant of the adjugate of a $3 \times 3$ matrix $M$ is given by $(\det M)^2$. Leveraging the property $\det(AB) = \det(A)\det(B)$, we can express the given equation $\det (\operatorname{adj} (2 A - A^{T}) \cdot \operatorname{adj} (A - 2 A^{T}) ) = 2^{8}$ as $(\det(2A - A^T))^2 \cdot (\det(A - 2A^T))^2 = 2^8$. Since the determinant of a matrix is equal to the determinant of its transpose, and $A^T - 2A$ is the transpose of $A - 2A^T$, we observe that $\det(A^T - 2A) = \det(A - 2A^T)$. By factoring out $-1$ from the term $2A - A^T$, which equals $-(A^T - 2A)$, we find that its determinant is $-\det(A^T - 2A) = -\det(A - 2A^T)$. Substituting these relationships, the equation simplifies to $(-\det(A - 2A^T))^2 \cdot (\det(A - 2A^T))^2 = (\det(A - 2A^T))^4 = 2^8$. This implies $(\det(A - 2A^T))^2 = 16$, or $|\det(A - 2A^T)| = 4$.

Constructing the matrix $A - 2A^T$ explicitly by subtracting twice the transpose of $A$ from $A$ yields:
$\begin{bmatrix} -1 & 0 & \alpha \\ -3 & 0 & -1 \\ -2\alpha & -1 & -2 \end{bmatrix}$
Calculating the determinant of this matrix by expanding along the second column results in $1 + 3\alpha$. Equating $|1 + 3\alpha| = 4$ and noting that $\alpha \in (0, \infty )$, we determine $\alpha = 1$. Substituting $\alpha = 1$ into matrix $A$, we obtain:
$A = \begin{bmatrix} 1 & 2 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{bmatrix}$
The determinant of $A$ is $1(0-1) - 2(2-0) + 1(1-0) = -1 - 4 + 1 = -4$. Consequently, $(\det(A))^2 = (-4)^2 = 16$.

A system of linear equations possesses infinitely many solutions when the equations are linearly dependent, meaning one equation can be expressed as a linear combination of the others, and the constant terms must remain consistent. Finding the relationship between the coefficients involves determining multipliers for the first two equations that generate the coefficients of the third. Multiplying $3x + 5y + \lambda z = 3$ by 9 and adding it to 10 times $7x + 11y - 9z = 2$ produces the combined equation $97x + 155y + (9\lambda - 90)z = 47$.

For the system to possess infinitely many solutions, this expression must be identical to the third equation, $97x + 155y - 189z = \mu$. Equating the coefficients of $z$ gives $9\lambda - 90 = -189$, which simplifies to $9\lambda = -99$ and results in $\lambda = -11$. Similarly, matching the constants establishes $\mu = 47$. Substituting these values into $\mu + 2\lambda$ yields $47 + 2(-11)$, which simplifies to 25.

A system of linear equations possesses infinitely many solutions if the determinant of the coefficient matrix, denoted as $D$, is zero, along with the determinants $D_1$, $D_2$, and $D_3$ derived by replacing the corresponding columns with the constant terms. We start by calculating $D$ using the matrix coefficients:

.
Expanding this expression simplifies to $\alpha \beta + 3 + 4\beta - 18 - 2 - 3\alpha$, which yields the condition $\alpha \beta - 3\alpha + 4\beta = 17$ once $D$ is set to zero.

To isolate $\beta$, we evaluate $D_2$ by replacing the second column of the coefficient matrix with the constants $-4, 5, 3$:

.
Simplifying the expression results in $5\beta - 9 + 8\beta - 36 - 9 = 13\beta - 54$. Setting $D_2 = 0$ leads directly to $\beta = \frac{54}{13}$. Substituting this value back into the relation $\alpha \beta - 3\alpha + 4\beta = 17$ allows us to solve for $\alpha$:
$\alpha \left(\frac{54}{13}\right) - 3\alpha + 4 \left(\frac{54}{13}\right) = 17$.
Multiplying the entire equation by 13 results in $54\alpha - 39\alpha + 216 = 221$, which simplifies to $15\alpha = 5$, yielding $\alpha = \frac{1}{3}$. Finally, calculating the required expression $12\alpha + 13\beta$ gives $12\left(\frac{1}{3}\right) + 13\left(\frac{54}{13}\right)$, which simplifies to $4 + 54 = 58$.

Since the matrix satisfies $AA^T = I$, it is an orthogonal matrix, implying the additional property $A^TA = I$. Expanding the squared binomials requires caution due to non-commutative matrix multiplication, resulting in $(A + A^T)^2 = A^2 + AA^T + A^TA + (A^T)^2 = A^2 + 2I + (A^T)^2$ and $(A - A^T)^2 = A^2 - AA^T - A^TA + (A^T)^2 = A^2 - 2I + (A^T)^2$. Summing these two expressions simplifies to $2A^2 + 2(A^T)^2$ because the identity terms cancel out. Finally, applying the leading coefficient $\frac{1}{2}A$ yields $A(A^2 + (A^T)^2)$, which expands to $A^3 + A(A^T)^2$; substituting $AA^T = I$ into the latter term confirms the final result as $A^3 + A^T$.

The determinant of the product of two matrices is equal to the product of their individual determinants, namely $\det(AB) = \det(A) \det(B)$. Because matrix $B$ is the cofactor matrix of $A$, its determinant follows the property $\det(\text{cof}(A)) = \det(A)^{n-1}$, where $n$ is the order of the matrix. For a $3 \times 3$ matrix, this simplifies to $\det(B) = \det(A)^2$, establishing the overall relationship $\det(AB) = \det(A)^3$.

By identifying the constants as $\alpha = 2$ and $\beta = 1$, the matrix $A$ is defined as

. Evaluating the determinant of $A$ gives $1(8-2) - 2(8+1) + 3(4+2)$, which simplifies to $6 - 18 + 18 = 6$. Raising this value to the power of three results in $216$.

Applying row transformations $R_1 \to R_1 - R_2$ and $R_2 \to R_2 - R_3$ simplifies the determinant to

. Expanding this determinant gives $(\alpha - a)(\beta - b)\gamma - (\alpha - a)b(c - \gamma) + a(b - \beta)(c - \gamma) = 0$. Dividing the entire equation by the non-zero product $(\alpha - a)(\beta - b)(\gamma - c)$ yields $\frac{\gamma}{\gamma - c} + \frac{b}{\beta - b} + \frac{a}{\alpha - a} = 0$, confirming the sum is equal to 0.

Calculating matrix powers often reveals a pattern or a scalar identity that simplifies the evaluation of higher exponents. Squaring $A$ yields

, and repeating this process leads to

and

, which is equal to $-27I$. Since $A^{12} = (A^{6})^{2} = (-27)^{2}I = 729I$, the matrix $A^{13}$ becomes

. The sum of the diagonal elements is $1458 + 729 = 2187$, which is equivalent to $3^{7}$, confirming $n=7$.

A system of linear equations possesses a unique solution only when the determinant of its coefficient matrix is non-zero. Calculating this determinant yields the expression $2\lambda^2 - \lambda - 1 = 0$, which indicates that the determinant is zero when $\lambda = 1$ or $\lambda = -\frac{1}{2}$. For any value of $\lambda$ other than these two, the system will have a unique solution. When $\lambda = 1$, testing the augmented matrix reveals that the system has infinitely many solutions if $\mu = 13$ and is inconsistent if $\mu \neq 13$. Because the condition claiming a unique solution for all $\lambda \neq 1$ with $\mu \neq 13$ ignores the case where $\lambda = -\frac{1}{2}$, which also forces the determinant to zero and results in an inconsistent system, that specific statement is not correct.

A system of linear equations possesses infinitely many solutions when the determinant of the coefficient matrix $D$ equals zero, alongside the determinants $D_1$, $D_2$, and $D_3$, which are generated by substituting the constant column into the original coefficient matrix. The parameter $\mu$ is identified by evaluating the determinant $D_3$ and setting it equal to zero:
$\begin{vmatrix} 2 & 7 & 3 \\ 3 & 2 & 4 \\ 1 & \mu & -1 \end{vmatrix} = 0$
Expanding this matrix yields $2(-2 - 4\mu) - 7(-3 - 4) + 3(3\mu - 2) = 0$, which simplifies to $-4 - 8\mu + 49 + 9\mu - 6 = 0$. Combining the terms results in $\mu + 39 = 0$, establishing that $\mu = -39$.

With $\mu$ known, the parameter $\lambda$ is determined by setting the determinant of the coefficient matrix $D$ to zero:
$\begin{vmatrix} 2 & 7 & \lambda \\ 3 & 2 & 5 \\ 1 & -39 & 32 \end{vmatrix} = 0$
The expansion provides $2(64 + 195) - 7(96 - 5) + \lambda(-117 - 2) = 0$, which simplifies to $518 - 637 - 119\lambda = 0$. Solving this equation gives $-119 - 119\lambda = 0$, meaning $\lambda = -1$. Calculating the difference $\lambda - \mu$ yields $-1 - (-39)$, which results in 38.

The linearity of determinants allows us to simplify expressions when multiple matrices share identical columns or rows. Since the second and third columns of $A_r$ remain invariant for any $r$, we can compute $2A_{10} - A_8$ by focusing entirely on the difference in the first column. Substituting the values $r=10$ and $r=8$ into the expression, we compute the difference between the first columns of the two determinants:

$2A_{10} - A_8 = \begin{vmatrix} 12 & 1 & \frac{n^2}{2}+\alpha \\ 24 & 2 & n^{2-}\beta \\ 34 & 3 & \frac{n(3n-1)}{2} \end{vmatrix}$

To reduce this, applying the column operation $C_1 \to C_1 - 12C_2$ transforms the first column, producing zeros in the first two rows and simplifying the expansion process:

$\begin{vmatrix} 0 & 1 & \frac{n^2}{2}+\alpha \\ 0 & 2 & n^{2-}\beta \\ -2 & 3 & \frac{n(3n-1)}{2} \end{vmatrix}$

Expanding along this first column, the determinant becomes $-2 \left( (n^{2-}\beta) - 2(\frac{n^2}{2}+\alpha) \right)$. Simplifying the terms inside the parentheses leads to $-2(n^2 - \beta - n^2 - 2\alpha)$, which results in the final value $4\alpha+2\beta$.

Cramer's Rule provides a systematic approach for solving this system of linear equations, where the variables $x$ and $y$ are calculated as the ratio of determinants of modified matrices to the determinant of the coefficient matrix $A$. The determinant of $A$ is $|A| = (2)(m) - (-5)(3) = 2m + 15$. By replacing the columns of $A$ with the constant vector $B$, the variables are determined to be

and

.

The condition that $x < 0$ and $y < 0$ constrains the possible values of $m$. The expression for $x$ is negative for $m \in (-15/2, 0)$, while the expression for $y$ is negative for $m \in (-15/2, 30)$. Since the system requires both variables to be negative simultaneously, the valid interval for $m$ is $(-15/2, 0)$, which defines the limits of integration as $a = -15/2$ and $b = 0$. Evaluating the integral $8 \int_{-15/2}^{0} (2m + 15) \, dm$ involves finding the antiderivative $m^2 + 15m$ and applying the fundamental theorem of calculus, yielding 8 \[0 - ((-15/2)^2 + 15(-15/2))]$ = 8 [0 - (225/4 - 225/2)] = 8 \times 225/4$, which simplifies to 450.

For a system of three linear equations to possess infinitely many solutions, the determinant of the coefficient matrix must be zero, which indicates that the equations are linearly dependent. Evaluating the determinant of the coefficients yields:
$\begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & 5 \\ 1 & 2 & m \end{vmatrix} = 1(5m - 10) - 1(2m - 5) + 1(4 - 5) = 3m - 6$
Setting this determinant to zero gives $3m - 6 = 0$, which results in $m = 2$. With the system now dependent, we must ensure consistency by setting the determinant of the augmented matrix-formed by replacing the third column with the constants 4, 17, and $n$-to zero. Examining this determinant:
$\begin{vmatrix} 1 & 1 & 4 \\ 2 & 5 & 17 \\ 1 & 2 & n \end{vmatrix} = 1(5n - 34) - 1(2n - 17) + 4(4 - 5) = 3n - 21$
Equating this to zero leads to $3n - 21 = 0$, so $n = 7$. Substituting these values into the expression $m^2 + n^2 - mn$, we calculate $2^2 + 7^2 - (2)(7) = 4 + 49 - 14$, which equals 39.

Because the linear system has a non-trivial solution, the matrix of coefficients must have a determinant of zero to ensure linear dependence among the vectors:
$\begin{vmatrix} \alpha & 1 & 2 \\ 1 & \beta & 3 \\ 2 & 2 & \gamma \end{vmatrix} = 0$
Expanding this determinant along the first row gives $\alpha(\beta\gamma - 6) - 1(\gamma - 6) + 2(2 - 2\beta) = 0$, which simplifies to the expression $\alpha\beta\gamma - 6\alpha - \gamma + 6 + 4 - 4\beta = 0$. Substituting the known value $\alpha\beta\gamma = 45$ into this equation results in $45 - 6\alpha - 4\beta - \gamma + 10 = 0$, and rearranging these terms leads to $6\alpha + 4\beta + \gamma = 55$.

The characteristic equation $\det(A - xI) = 0$ provides the eigenvalues of a matrix $A$. Since the roots of this equation are -1 and 3, the eigenvalues of $A$ are $\lambda_1 = -1$ and $\lambda_2 = 3$. According to the spectral mapping theorem, the eigenvalues of the matrix $A^2$ are the squares of the eigenvalues of $A$, which are $(-1)^2 = 1$ and $3^2 = 9$. The trace of any square matrix is equal to the sum of its eigenvalues, so the sum of the diagonal elements of $A^2$ is calculated as $1 + 9 = 10$.